-4.905t^2+29.4t=0

Solution for -4.905t^2+29.4t=0 equation:

-4.905t^2+29.4t=0

a = -4.905; b = 29.4; c = 0;
Δ = b2-4ac
Δ = 29.42-4·(-4.905)·0
Δ = 864.36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29.4)-\sqrt{864.36}}{2*-4.905}=\frac{-29.4-\sqrt{864.36}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29.4)+\sqrt{864.36}}{2*-4.905}=\frac{-29.4+\sqrt{864.36}}{-9.81}
$